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ZanzabumX [31]
4 years ago
11

Evaluate P(6, 6). 720 01 6

Mathematics
1 answer:
Vinvika [58]4 years ago
6 0
The correct answer is 720 I did this problem long time ago hope this helps :)

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Graph for f(x)=6^6 and f(x)=14^x
zlopas [31]

Graph Transformations

There are many times when you’ll know very well what the graph of a

particular function looks like, and you’ll want to know what the graph of a

very similar function looks like. In this chapter, we’ll discuss some ways to

draw graphs in these circumstances.

Transformations “after” the original function

Suppose you know what the graph of a function f(x) looks like. Suppose

d 2 R is some number that is greater than 0, and you are asked to graph the

function f(x) + d. The graph of the new function is easy to describe: just

take every point in the graph of f(x), and move it up a distance of d. That

is, if (a, b) is a point in the graph of f(x), then (a, b + d) is a point in the

graph of f(x) + d.

As an explanation for what’s written above: If (a, b) is a point in the graph

of f(x), then that means f(a) = b. Hence, f(a) + d = b + d, which is to say

that (a, b + d) is a point in the graph of f(x) + d.

The chart on the next page describes how to use the graph of f(x) to create

the graph of some similar functions. Throughout the chart, d > 0, c > 1, and

(a, b) is a point in the graph of f(x).

Notice that all of the “new functions” in the chart di↵er from f(x) by some

algebraic manipulation that happens after f plays its part as a function. For

example, first you put x into the function, then f(x) is what comes out. The

function has done its job. Only after f has done its job do you add d to get

the new function f(x) + d. 67Because all of the algebraic transformations occur after the function does

its job, all of the changes to points in the second column of the chart occur

in the second coordinate. Thus, all the changes in the graphs occur in the

vertical measurements of the graph.

New How points in graph of f(x) visual e↵ect

function become points of new graph

f(x) + d (a, b) 7! (a, b + d) shift up by d

f(x) Transformations before and after the original function

As long as there is only one type of operation involved “inside the function”

– either multiplication or addition – and only one type of operation involved

“outside of the function” – either multiplication or addition – you can apply

the rules from the two charts on page 68 and 70 to transform the graph of a

function.

Examples.

• Let’s look at the function • The graph of 2g(3x) is obtained from the graph of g(x) by shrinking

the horizontal coordinate by 1

3, and stretching the vertical coordinate by 2.

(You’d get the same answer here if you reversed the order of the transfor-

mations and stretched vertically by 2 before shrinking horizontally by 1

3. The

order isn’t important.)

74

7:—

(x) 4,

7c’

‘I

II

‘I’

-I

5 0
3 years ago
Which of the following explains why f(x) = log4x<br> does not have a y-intercept?
Margarita [4]

Answer: f(x) approaches infinity

Step-by-step explanation:

N/A

7 0
3 years ago
What is the minimum and maximum of f(x)=-10x^3+6x-2
const2013 [10]

Answer:

Minimum: (\frac{\sqrt{5} }{5},-2+\frac{4\sqrt{5}  }{5})

Maximum: (-\frac{\sqrt{5} }{5} ,-2-\frac{4\sqrt{5} }{5} )

Step-by-step explanation:

5 0
3 years ago
Suppose it takes 4 hours for a certain strain of bacteria to reproduce by dividing in half. If 70 bacteria are present to begin
Sauron [17]

Answer:

1 day: 4480

2 days: 8960

3 days: 13440

f^{-1}(x) = \sqrt{\frac{x -4}{4}} ---- function inverse

Step-by-step explanation:

Given

f(x) = 70 \cdot 64x

Solving (a): The amount present after 1 day.

Here, x =1

So:

f(1) = 70 \cdot 64*1= 4480

Solving (b): The amount present after 2 days.

Here, x =2

So:

f(2) = 70 \cdot 64*2= 8960

Solving (c): The amount present after 3 days.

Here, x = 3

So:

f(3) = 70 \cdot 64*2= 13440

Solving (d): The inverse function of:

f(x)= 4x^2 + 4

Replace f(x) with y

y= 4x^2 + 4

Swap x and y

x= 4y^2 + 4

Rewrite as:

4y^2 = x -4

Divide by 4

y^2 = \frac{x -4}{4}

Take square roots

y = \sqrt{\frac{x -4}{4}}

Replace y with function inverse

f^{-1}(x) = \sqrt{\frac{x -4}{4}}

6 0
3 years ago
3(-n+4)+5n=2n <br> a. n=3<br> b. no solution<br> c. infinitely many solutions
777dan777 [17]

Answer:

no solutions

Step-by-step explanation:

3(-n+4)+5n=2n

Distribute

-3n +12 +5n = 2n

2n+12 = 2n

Subtract 2n from each side

12 = 0

This is never true so there are no solutions

4 0
3 years ago
Read 2 more answers
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